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MHT CET 2023 (12 May, Shift 2) · Energy in SHM

Energy in SHM question from MHT CET 2023 (12 May, Shift 2)

Q46MHT CET 2023 · 12 MAY · SHIFT 2Medium

Energy in SHM

A particle starts from mean position and performs S.H.M. with period 4 second. At what time its kinetic energy is 50%50\% of total energy?

(cos⁡45∘=12)\left(\cos 45^\circ = \frac{1}{\sqrt{2}}\right)

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Answer: (d)

  1. K=50%K = 50\% of EE at x=A2x = \frac{A}{\sqrt{2}}, so sin⁡ωt=12⇒ωt=π4\sin\omega t = \frac{1}{\sqrt{2}} \Rightarrow \omega t = \frac{\pi}{4}.
  2. t=T8=48=0.5t = \frac{T}{8} = \frac{4}{8} = 0.5 s
  3. So the answer is (d) 0.50.5 s.
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